A first order reaction has a rate constant of $2.303\times 10^{-3}{\text{s}}^{-1}$. The time required for $40\text{g}$ of this reactant to reduce to $10\text{g}$ will be (Given ${\log }_{10}2=0.3010$):

Solution & Explanation

\textbf{Answer:} (D) 602 s \textbf{Solution:} For a first order reaction, $k=2.303\times 10^{-3}{\text{s}}^{-1}$. The half-life is $t_{1/2}=\frac{0.693}{k}=\frac{0.693}{2.303\times 10^{-3}}=301\text{s}$. Reducing $40\text{g}\to 10\text{g}$ is two successive halvings ($40\to 20\to 10$), i.e. 2 half-lives. $t=2\times t_{1/2}=2\times 301=602\text{s}$ Equivalently: $t=\frac{2.303}{k}\log \frac{40}{10}=\frac{2.303}{2.303\times 10^{-3}}\times \log 4=1000\times 2\times 0.3010=602\text{s}$.